Utilizing the Oberth effect

[Kerbrud]

Hi all!

In my preparation for interplanetary travel, I thought I needed to examine the Oberth effect a bit, in order to spend the least amount of fuel. The thesis was that if I incrementally gained my target altitude, I'd end up using less fuel than reaching the target altitude in one go. The results, below, puzzles me a bit:

Target altitude: Ap = 2 000 000 m

Engine used = LV-909

[table=width: 1100]

[tr]

[td]Ap = 302 km[/td]

[td]Ap = 802 km[/td]

[td]Ap = 1510 km[/td]

[/tr]

[tr]

[td]Pe = 151 km[/td]

[td]Pe = 152 km[/td]

[td]Pe = 152 km[/td]

[/tr]

[tr]

[td]Speed @ Pe = 2265.7 m/s[/td]

[td]Speed @ Pe = 2472.0 m/s[/td]

[td]Speed @ Pe = 2629.0 m/s[/td]

[/tr]

[tr]

[td]Delta V required to target = 436.5 m/s[/td]

[td]Delta V required to target = 227.3 m/s[/td]

[td]Delta V required to target = 68.1 m/s[/td]

[/tr]

[tr]

[td]Delta V required to next Ap (802 km) = 209.0 m/s [/td]

[td]Delta V required to next Ap (1510 km) = 158.9 m/s[/td]

[td]Delta V required to next Ap (2000 km) = 68.1 m/s[/td]

[/tr]

[/table]

As you can see, the required delta V (302->2000) = 436.5 m/s = required delta V (302->802->1510->2000)

Ergo, my thesis failed. Does anyone care to explain to me why it failed, and why I should care about the Oberth effect at all? It's clear that the higher your Ap is, the less delta V is required to gain it at Pe, but it didn't help me a bit in this experiment.

According to the Wikipedia page on the subject, http://en.wikipedia.org/wiki/Oberth_effect, the best results are achieved when using engines with a lot of power; giving a lot of force in a short time. This wasn't the case with the craft I made (Orion replica). This could be the reason, but I would think that using a less efficient engine would only result in the Oberth effect being less efficient - not being totally non-existing.

An example from the real world, here's India's Mars probe gaining altitude incrementally:

Edited January 3, 2014 by Kerbrud

[Kasuha]

Oberth effect near Kerbin means, for instance, that if you burn to just eject from Kerbin SOI and then, when in interplanetary space, burn to raise your orbit e.g. to Jool, you spend about 1/3 more dv than if you both eject and raise your orbit to Jool right from Kerbin LKO.

The reason for gradual increasing your apoapsis is that if your burn would take too long, you'd get too far away from Kerbin/Earth and the burn would lose its effectiveness. You can't really simulate it with maneuver nodes because these assume applying whole dv amount at the exact point where they are placed.

If your rocket has enough thrust, there's no reason to raise the orbit gradually. You can do it in single burn.

[Kerbrud]

Oberth effect near Kerbin means, for instance, that if you burn to just eject from Kerbin SOI and then, when in interplanetary space, burn to raise your orbit e.g. to Jool, you spend about 1/3 more dv than if you both eject and raise your orbit to Jool right from Kerbin LKO.

The reason for gradual increasing your apoapsis is that if your burn would take too long, you'd get too far away from Kerbin/Earth and the burn would lose its effectiveness. You can't really simulate it with maneuver nodes because these assume applying whole dv amount at the exact point where they are placed.

If your rocket has enough thrust, there's no reason to raise the orbit gradually. You can do it in single burn.

Your first paragraph seemed a bit unclear, but you mean that going directly from the launchpad to Jool uses 1/3 more fuel than first gaining LKO, then raise the orbit to Jool?

About simulating using the maneuver-nodes: Although they imply reaching target speed at one point, you end up using pretty much the same amount of delta V if you burn at (time to node = 0) - (est. burn/2). The difference between the two lies in how accurate to your desired orbit you'll end up going. The importance of this increases, as you say, the longer the burn takes. I guess the effectiveness lies in the amount of correction burns you'd have to do later - which I guess would be the reason for India's probe gaining altitude incrementally.

Please answer this post's first question, then I have a follow-up=P

[Kasuha]

Your first paragraph seemed a bit unclear, but you mean that going directly from the launchpad to Jool uses 1/3 more fuel than first gaining LKO, then raise the orbit to Jool?

No.

Starting on LKO, you can either perform a maneuver which will get you on an intercept trajectory with your target, or you can just raise your apoapsis till it exits Kerbin LKO which will put you on interplanetary trajectory very close to Kerbin's. You can then prepare the intercept maneuver towards your target from there, but it will cost you more dv because you'll lose the Oberth effect of Kerbin.

[Kerbrud]

Right. Then I've been doing the most efficient thing, actually.

Thanks!

[Aphobius]

lose its effectiveness

Or this from wikipedia

In astronautics, the Oberth effect is where the use of a rocket engine when travelling at high speed generates more useful energy than one at low speed. The Oberth effect occurs because the propellant has more usable energy due to its kinetic energy on top of its chemical potential energy. The vehicle is able to employ this kinetic energy to generate more mechanical power.

These are explanations given by people that are not understanding what really happens. I really don't get that wikipedia explanation ... it's like ... wtf?

Simple explanation (makes me look like an idiot ... but everyone should be able to understand this, unlike explanations from wikipedia or other people that appear to be smart, but no one is understanding what they say):

Let's say you have a planet without an atmosphere, you burn straight up and you want to leave SOI with an X speed.

If you get the required speed instantly after the launch, you'll leave the SOI fast.

If you have a weak rocket and let's say you have to burn all the way from the launch until you leave SOI, to be able to leave it with the same X speed ... you'll leave SOI slower.

The required burn is the final speed + gravitational acceleration over time (which is also a speed ... acceleration over time is a speed).

In the first case final speed is X, but gravitational acceleration over time is smaller (because time is shorter).

In the second case final speed is also X, but gravitational acceleration over time is larger (because time is longer).

With orbits it gets a little more complicated, but it's basically the same thing.

Your experiment failed because the Oberth effect is not involved if you don't leave SOI. You haven't chosen your "objective" well.

"Target altitude: Ap = 2 000 000 m" is not good for analyzing the Oberth effect, if you want to observe the Oberth effect, your objective must be the speed at the moment you leave SOI.

Pretty hard to make an experiment like this in KSP.

EDIT: The reason for which that thing from your picture is working is because if you think about the Pe as the launchpad from my earlier example. After each Ap raising maneuver, the next time you get to the Pe, it's like being on the "launchpad" with a higher speed.

So in your experiment, if you make a single burn, you start from the "launchpad" with 2265.7 m/s, but if you do that 2 maneuvers, you start from the "launchpad" with 2629.0 m/s.

So it's like you've got that ~400m/s instantly, just after the "launch".

Edited January 3, 2014 by Aphobius

[Kerbrud]

The required burn is the final speed + gravitational acceleration over time (which is also a speed ... acceleration over time is a speed).

In the first case final speed is X, but gravitational acceleration over time is smaller (because time is shorter).

In the second case final speed is also X, but gravitational acceleration over time is larger (because time is longer).

I think you got your equations wrong.

First case:

required delta V = final speed + gravitational acceleration over time = final speed + acceleration/short time = final speed + more additional speed

Second case:

required delta V = final speed + acceleration/long time = final speed + less additional speed

This would mean that the shorter the time, the larger the delta V would become. In addition you'd have to concider that the ratio acc/time is probably heavily weighted towards the acceleration of the chosen engine.

So, I do get that it's happening, but I'm still not convinced as to why.

I see why my experiment failed, though. Although it doesn't save any fuel (other than in the manner of accuracy which I explained further up), it's interesting to see that in order to maintain a larger apoapse, the speed at the periapse increases by the second degree in relation with the altitude of the apoapse. Because of this, the reqired delta V to further gain the apoapse decreases also in the second degree.

[Kerbrud]

So, this takes care of the orbital section of the Oberth effect, at least!=)

But i'm still at a loss when it comes to why your first case is explaining the oberth effect at all, other than the additional speed you get from Kerbin's speed revolving around Kerbol.

I may be stupid as heck here, I'm sorry=P

[-Velocity-]

To understand the Olberth effect, think of it this way. If you've had at least high school level physics, remember the equation:

Work = Force*Distance

Work is the amount of energy added or subtracted from the object that is moving.

This equation is actually a simplification. In actuality, it's

Work = Force (dot) Distance,

where (dot) is the vector dot product. For simplicity's sake though, since your physics is probably not very good, let's just simplify this to one dimension, so that the force can only be applied in the direction of motion, and in the opposite direction of motion. When the force is applied in the same direction as the motion, then the work done is positive, and if the force is applied in the opposite direction as the motion, the work done is negative.

Anyway, just LOOK at the equation. If you want to maximize the amount of work done, you want to be moving over as much distance as possible while the force is being applied. In other words, when applying a force for a fixed amount of time, you do more work if you are moving faster because the distance over which you apply the force is greater.

Another way to show the same thing with high school physics is to consider the equation for kinetic energy,

KE = 1/2*mass*velocity^2

Say you have a spacecraft that can change its delta-V by 100 m/s. If that delta-V is applied when the spacecraft is initially moving 1000 m/s, you get an increase in kinetic energy of

1/2*((1000 + 100)^2 - 1000^2) = 105 kJ/kg.

BUT, if you apply that delta-V when the spacecraft is initially going 2000 m/s, you get an increase in kinetic energy of

1/2*((2000 + 100)^2 - 2000^2) = 205 kJ/kg.

Edited January 3, 2014 by |Velocity|

[Aphobius]

I'm a grumpy person and I don't like people to tell me that I'm wrong.

I'll explain again what you've understood wrong and after that just read again my first post.

Let's say short time means 10 seconds and long time 50 seconds. If gravitational acceleration is 10m/s^2, in 10 seconds that means 100m/s, in 50 seconds that means 500m/s.

I don't know how you've got this wrong. But in case it still isn't clear enough:

-planet without atmosphere

-gravitational acceleration 10m/s^2

-planet radius = 500km

-SOI radius = 2500km (stupid number that can't happen in real life, but it's just an example so the calculation are easier to understand)

If you have a really powerful rocket and you get 6325m/s instantly after the launch, you'll leave SOI with 1m/s after 625 seconds. (this is stupid, it doesn't take into consideration that gravitation gets weaker with the distance, but that would be too complex for this post and that calculation is not really required for this example).

Now think that you have some crap rocket that got a TWR of 1. Kerbals start the rocket and they give it a push with their hands so it gets to 1m/s. After that, the rocket will maintain that 1m/s (because it got a TWR of 1). You will leave the SOI in 2,000,000 seconds and you will need a delta-v of 20,000,000m/s .... ok, 20,000,001m/s (including the push kerbals gave it with their hands).

Now I hope this part is clear enough, you can read my first post again.

EDIT: I can add something to explain Oberth effect even better, in that example with 6325m/s, if you do that and you leave the SOI with 1m/s, if you burn another 6325m/s after leaving the SOI, you'll have that ... 6326m/s ... but if you burn that 6325m/s from the launch and your speed will be 12650m/s just after the launch, you are going to leave the SOI after 169 seconds with a speed of 11km/s.

So with the same amount of fuel, if you do only one burn at the launch (12650m/s) you leave SOI with 11km/s.

But if split the burn in 2, one at the launch (6325m/s) and one after you leave the SOI (also 6325m/s) you are leaving the SOI with only 6326m/s.

EDIT 2: To check my calculations you can use the normal equations of motion, with distance = 2000km and acceleration = -10m/s^2. This doesn't take into consideration that the gravitation is getting weaker with the distance, but that's not important for this explanation (the numbers would be different, but the ratios would be the same).

Edited January 3, 2014 by Aphobius

[nholzric]

I think you got your equations wrong.

First case:

required delta V = final speed + gravitational acceleration over time = final speed + acceleration/short time = final speed + more additional speed

Second case:

required delta V = final speed + acceleration/long time = final speed + less additional speed

When Aphobius says "over time", he does not mean "divided by time" he means "applied over a period of time". The less time you spend letting gravity slow you down, the less initial speed you need.

I don't understand the rest of Aphobius's explanation or how it explains the Oberth effect, but you misinterpreted these equations.

[Aphobius]

When Aphobius says "over time", he does not mean "divided by time" he means "applied over a period of time". The less time you spend letting gravity slow you down, the less initial speed you need.

I don't understand the rest of Aphobius's explanation or how it explains the Oberth effect, but you misinterpreted these equations.

Yeah, that's what I've meant.

If you don't understand the first post, you can read the second one too, I don't know if you've seen it.

Oberth effect is explained in a stupid way, that crap with energy etc. That thing works, you can calculate stuff and it will approximate real life pretty well, but I doubt there are too many people that really understand that mass * velocity, even physicist, they know the equations, but it's not like they understand anything from them, they just replace things in them and they get the correct answer.

I think my explanation is clear enough for everyone, using the Oberth effect means spending as least as possible time in the gravity of a planet, so the gravity of the planet acts on you a shorter time. Not that crap with energy etc.

EDIT: I was not thinking about that powered flyby thing, yes, Oberth effect is not only about spending as least as possible time in the gravity of a planet.

You can also do that thing, to enter the SOI of a planet and make a burn at the Pe, you use little delta-v to carry out of the planets SOI a large amount of the speed you've got from the gravitational acceleration while you were falling towards Pe.

Actually I'm an idiot =)) =)) The explanation can't get simpler than that thing with the energy if you want it to be complete. But as I've said, that thing is pretty hard for most of the people to understand and my I guess if someone is not understanding Oberth effect at all, can still learn something from my explanation.

Edited January 4, 2014 by Aphobius

[CombatWombat]

I have no idea what you are trying to explain. Gravity losses maybe, but not Oberth.

We know Power = work / time.

and work = force*distance.

and velocity is distance / time.

Therefor we can do some substitution and see that Power = Force*Velocity

The force generated by a rocket doesn't really give a crap about it's velocity (unlike a car or whatever we are normally more accustomed to).

So taking the force to be constant, it's clear that the power generated by the rocket is greater when it is operating at a higher velocity.

The bit I don't get, and where I think a lot of people get confused... Velocity is a relative quantity.

If my target is Duna, what velocity do I "maximize"? I have a velocity relative Kerbin. A velocity relative to Duna. A velocity relative to Kerbol. And a velocity relative to some small furry creatures from Alpha Centuri.

I would suspect the answer is "Velocity Relative Kerbol (Sun)", but I can't explain why.

I made a diagram to explain my confusion...

Shown is an orbit equal to minmus (A). And an orbit *around* minmus (B,C).

We can burn at A, B, or C to go to...someplace. Duna. Whatever.

Let's ignore the fact that some of them take you off in entirely the wrong direction...

C sounds like the winner, but I guess it took fuel to GET to that position to start with?

Do I even *care* about my velocity relative to Kerbin, if I'm interested in a cheap orbit around SUN?

[Kerbrud]

I'm a grumpy person and I don't like people to tell me that I'm wrong.

Based on your exact wording, I somehow (stupidly) managed to claim that v = v/t. Your wording was acceleration over time, which would be a/t. But, had you said acceleration TIMES time (which I of course should've known to correct myself) I would've followed you. This may be due to a language incompatibility (in Norway, a/b would be "a over b").

Unless I messed up something again, with the powerful rocket you'd leave SOI after 624 s.

[Aphobius]

I have no idea what you are trying to explain. Gravity losses maybe, but not Oberth.

Yes, that's Oberth effect, it can be described as that ... gravitational losses. Again, this explanation is stupid and with simple words but you should be able to understand the Oberth effect from it, if you can't understand the scientific explanation.

Practically Oberth effect can be used in 2 situations while flying spaceships.

1. Launch

You want to accelerate as fast as possible, to reduce the gravitational losses. You spend less time in planets gravity, you lose less velocity because of it.

2. Powered fly-by

If you enter the SOI of a planet, being on a hyperbolic orbit (almost line because you go very fast), you enter the SOI with a speed and you leave the SOI with the same speed (relative to that planet).

But if you make a burn at the Pe, if that burn is going to be a 1000m/s burn, when you leave the SOI of the planet, instead of leaving it with the same speed you've entered it, you'll leave it NOT with entry speed + 1000m/s, you'll leave it with entry speed + 1000m/s + something more because of the Oberth effect.

This is not hard to understand.

Let's say you have 500m/s when you enter SOI and 3000m/s at Pe. So from Pe to the point where you leave the SOI, you lose 2500m/s in X seconds (let's say 100 seconds).

Now if you make a 1000m/s burn at Pe, someone that doesn't know about Oberth effect would think that you leave SOI with 1500m/s, but this would mean that from 4000m/s (3000m/s that you would have anyway + 1000m/s from the burn) you would have to loose 2500m/s to get to 1500m/s.

We know it takes us 100 seconds to lose that 2500m/s, but that 100 seconds are with a starting velocity of 3000m/s.

If our starting velocity is 4000m/s, we'll leave SOI in less than 100 seconds, so we'll lose less than 2500m/s.

Resulting that after the 1000m/s burn we've made at Pe, we'll leave SOI with a speed around 2000m/s (too lazy to calculate the exact value, you get what I mean anyway).

So that 1000m/s burn at the Pe resulted in a 1500m/s speed increase after we left the SOI.

PS: Kerbrud, I have another comment on the first page, is the last one, I think you didn't saw it.

EDIT: CombatWombat, yes, you care about your velocity around Kerbin, if you read carefully what I've written above and you understand it, you'll understand that you can carry a part of your velocity around Kerbin out of Kerbin's SOI.

Edited January 4, 2014 by Aphobius

[Kerbrud]

KE = 1/2*mass*velocity^2

Yeah, I actually didn't remember that one, but it would only explain why you would need less delta V (because you'd be going faster, thus requiring less kinetic energy). Anyways, thanks for reminding me, it made explaining the failed experiment easier.

KE = 1/2*mass*velocity^2

Work = Force (dot) Distance,

where (dot) is the vector dot product. For simplicity's sake though, since your physics is probably not very good, let's just simplify this to one dimension, so that the force can only be applied in the direction of motion, and in the opposite direction of motion.

Well, at least the simplification is true in a circular orbit. If the engine is powerful enough that it can reach a gained orbit over a very little timespan, then certainly, force could be applied only along the direction of motion. If the orbit were elliptical, then yes, you'd have to consider the other vector components as well. But calculating that should be a simple manner of trigonometry.

The main issues in this thread were that I didn't understand it initially (you forget a lot of stuff way too fast when you don't use it every day, like in school), and something so simple as misinterpreting and writing error.

Oh well!

Time to blow stuff up!

[Aphobius]

About my post above, when I was saying launch, I was referring to a direct ascent. I have to admit, you can't really talk about the Oberth effect in that case.

The second example with the powered flyby was fine.

The first example should have been about making a transfer burn to other planet from an orbit around a planet. You want to do all the acceleration required for the transfer near the Pe (and you want Pe to be as low as possible), to carry a part of your speed around the planet out of it's SOI.

[NikkyD]

Aphobius is explaining gravity assists and NOT oberth.

Oberth in simple means, u change ur orbit the most (stretch ur AP) if you burn in the PE.

Pick some orbit, create a node with 500 m/s prograde. Now circle this node around the orbit (it should be a non circular orbit). In the AP the resulting orbit has the least distance traveled and in the PE it has the most distance traveled (because of most gain in kinetic energy and stuff).

The whole effect simply describes what changes ur orbit the most.

[Aphobius]

Aphobius is explaining gravity assists and NOT oberth.

Oberth in simple means, u change ur orbit the most (stretch ur AP) if you burn in the PE.

Pick some orbit, create a node with 500 m/s prograde. Now circle this node around the orbit (it should be a non circular orbit). In the AP the resulting orbit has the least distance traveled and in the PE it has the most distance traveled (because of most gain in kinetic energy and stuff).

The whole effect simply describes what changes ur orbit the most.

I'm not explaining gravity assists, if you claim you understand Oberth effect, read again what I've wrote.

Gravity assist is increasing your speed relative to the Sun, without you making a burn.

You enter the SOI of a planet from the retrograde direction of that planet, you leave the SOI of that planet in the prograde direction with a speed around the Sun = to the speed that you had when you entered the SOI + the speed of that planet. This is a gravity assist.

What I've explained, with the burn at Pe, was the Oberth effect.

Edited January 4, 2014 by Aphobius

[NikkyD]

2. Powered fly-by

If you enter the SOI of a planet, being on a hyperbolic orbit (almost line because you go very fast), you enter the SOI with a speed and you leave the SOI with the same speed (relative to that planet).

But if you make a burn at the Pe, if that burn is going to be a 1000m/s burn, when you leave the SOI of the planet, instead of leaving it with the same speed you've entered it, you'll leave it NOT with entry speed + 1000m/s, you'll leave it with entry speed + 1000m/s + something more because of the Oberth effect.

This is not hard to understand.

This is wrong. Oberth effect cannot generate velocity out of nothing, if you burn 1000 m/s delta_v, you will have 1000 m/s more v at the end. If you use a gravity assist, then u can indeed produce velocity out of "nothing" but that is not oberth...

[Kasuha]

This is wrong. Oberth effect cannot generate velocity out of nothing, if you burn 1000 m/s delta_v, you will have 1000 m/s more v at the end. If you use a gravity assist, then u can indeed produce velocity out of "nothing" but that is not oberth...

Actually no, Aphobius is right.

[SnappingTurtle]

The Oberth effect doesn't give you energy for free, it's just the fact that you get more energy for the same impulse at a higher velocity.

[LethalDose]

This is wrong. Oberth effect cannot generate velocity out of nothing, if you burn 1000 m/s delta_v, you will have 1000 m/s more v at the end. If you use a gravity assist, then u can indeed produce velocity out of "nothing" but that is not oberth...

The Oberth effect isn't creating something from nothing, and it's really not about delta-V. It's about getting the most orbital energy out of the delta-v you spend. The orbital energy is what's used to get to reach or escape SOI's, and the goal is to perform these maneuvers for as little delta-V as possible.

The Oberth effect is about getting more orbital energy for less delta-V, specifically taking advantage of the equation for kinetic energy: KE = 1/2 mv^2

I think someone linked

by Scott Manley demonstrates this extremely well. One exit maneuver w/o using the oberth effect runs out of fuel short of it's target, while the second maneuver with an oberth effect gravity assist (dropping Pe to just above the atmosphere and burning near Pe) not only suceeds, but leaves almost a half tank of fuel remaining.

And I actually just had a massive OH! moment writing this post RE: why Minmus refueling stations are FREAKING GENIUS for interplanetary missions. I must build one!

Since the velocity term is squared, the amount of kinetic energy gained by a fixed amount of delta-v varies by the initial orbital velocity. |Velocity| did a great job of demonstrating this above. It's a place where you can look at the equation and understand the implications. I really don't get the opinion expressed above reducing the value of scientists and equation. "Ignorant" would be the best possible word for it.

[SRV Ron]

From the Kerbal school of hard knocks. another post on a similar subject;

Two stage stock part rocket to determine the most efficient way to reach Duna orbit with this design.

1. Straight up dawn launch to Duna orbit

2. Launch to 100+k orbit, do an escape burn, do the Duna intercept burn. (Most people use this method to achieve interplanetary intercept.)

3. Launch to 100+k orbit, then do an escape to direct Duna intercept.

The rocket used has a stock capsule, parachute, decoupler, FL-T800 fuel can, LV-N engine, decoupler. Booster stage is a single Orange fuel can and Skipper engine. Ox Stat solar panels were added to keep the capsule battery charged.

The results.

1. A straight up burn at dawn to Duna intercept left 162 units of fuel left in the FL-T800 fuel can.

2. From the 100+K orbit, it took about 913 delta V to Kerbal escape, then another 892 delta V to reach Duna intercept for a total of 1,805 delta V. That left 154 units of fuel in the FL-T800 fuel can.

3. From the 100+K orbit to direct Duna intercept only took 1165 delta V leaving 215 units of fuel in the FL-T800 fuel can, a savings of 640 delta V or 61 units of fuel for this design.

And, a bit of a surprise that a straight up launch happened to be slightly more efficient then the three step orbit, escape, Duna burn. The most efficient, the one from orbit to Duna intercept.

[technion]

The frustrating part:

Using the previous post to create a highly efficient Duna intercept using "method 3" taking 1165 dv, followed swiftly by spending 1800dv correcting an inclination that's almost 90 degrees to ike (which I was heading for)